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Michele_Laino
 one year ago
Tutorial on the Quaternions of Hamilton
Michele_Laino
 one year ago
Tutorial on the Quaternions of Hamilton

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Thank you Michele, I will give it a read! :)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.8thanks!! :) @Astrophysics

Oleg3321
 one year ago
Best ResponseYou've already chosen the best response.0is this some sort of new language??? lol

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.8no, the quaternions were introduced by W. R. Hamilton in the far 1833 :) lol

Oleg3321
 one year ago
Best ResponseYou've already chosen the best response.0lol looks very complicated

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.8no, they are not complicated, please study my definitions first @oleg3321

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Very easy to understand, thanks again!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.8:) @Astrophysics

welshfella
 one year ago
Best ResponseYou've already chosen the best response.0Interesting article. Well explained. Hamilton was Ireland's most famous mathematician . Unfortunately his life was cut short by alcoholism.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.8Thanks! @welshfella

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0I remember my calc 3 teacher mentioning those to look at if you got bored.. thanks @Michele_Laino

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.8thanks! :) @DanJS

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0Thank you @Michele_Laino . I will definitely give it a read

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.8Thanks! :) @arindameducationusc

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0quaternions aren't a field, they are a division ring

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.8Quaternions are a field @oldrin.bataku

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0also you should introduce those other basis matrices \(\mathbf i,\mathbf k\) more naturally  something like this: consider for \(z=a+bi\) we have: $$\begin{align*}a+bi\mapsto \begin{pmatrix}a+bi&0\\0&abi\end{pmatrix}&=\begin{pmatrix}a&0\\0&a\end{pmatrix}+\begin{pmatrix}bi&0\\0&bi\end{pmatrix}\\&=a\begin{pmatrix}1&0\\0&1\end{pmatrix}+b\begin{pmatrix}i&0\\0&i\end{pmatrix}\\&=a\cdot\mathbf1+b\cdot\mathbf i\end{align*}$$ and then we have that \(\mathbf i\cdot\mathbf j=\mathbf k\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0https://en.wikipedia.org/wiki/Quaternion#Noncommutativity_of_multiplication >The fact that quaternion multiplication is not commutative makes the quaternions an oftencited example of a strictly skew field.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.8I wrote that formula: \[{\mathbf{ij}} =  {\mathbf{ji}} = {\mathbf{k}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0https://en.wikipedia.org/wiki/Division_ring >In abstract algebra, a division ring, also called a skew field, is a ring in which division is possible. >Division rings differ from fields only in that their multiplication is not required to be commutative. ... Historically, division rings were sometimes referred to as fields, while fields were called “commutative fields”.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, I know you did, but you introduced \(\mathbf k\) as a matrix first without any motivation. the reason for its existence is that we have four linearly independent matrices \(\mathbf {1,i,j,ij}\) since the multiplication is bilinear, and then we just call \(\mathbf {ij}=\mathbf k\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.8I never worked with noncommutative algebra

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.8I meant I never cited noncommutative algebra

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0? fields have commutative multiplication by definition, whereas the quaternions \(\mathbb H\) are a noncommutative algebra over \(\mathbb R\) and a division ring, not a field

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.8when I wrote a matrix in M2(C) it is supposed that the multiplication is noncommutative since as you know, the multiplication betwen matrices is noncommutative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol, yes, clearly, and these matrices do not commute, so they cannot form a field. the quaternions are not a field.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0For real man, tuts that are wrong are even worse. This is all simple definition stuff you are getting wrong...Please remove.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.8More precisely, I have used the term "field", since it is commonly used in mathematics literature, a more appropriate term can be "sfield" which indicate a noncommutative field. I have supposed that, being the noncommutative characteristic evident, there was no need to use the term "sfield"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0field is not commonly used in mathematics literature for a skew field, this isn't the 1800s anymore

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it's called a division ring, that is the appropriate, correct, and commonly used term for algebraic structures like this
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